2025 AMC 12A Problems/Problem 25: Difference between revisions

Revision as of 02:24, 7 November 2025

We begin by analyzing the sign chart. From the condition $f(x) \le 0$ on $[a,b] \cup (c,d)$ and $f(x) > 0$ elsewhere, we deduce the following:

- $a$ and $b$ are zeros of $f$ , since we transition from $f>0$ to $f\le0$ at $a$ , and from $f\le0$ to $f>0$ at $b$ . - $c$ and $d$ are poles or holes of $f$ , since on $(c,d)$ we have $f\le0$ , while immediately outside that interval $f>0$ .

Thus, the sign pattern of $f(x)$ is $\begin{array}{cccccccccc} & (-\infty, a) & a & (a,b) & b & (b,c) & c & (c,d) & d & (d,\infty) \\ f(x) & + & 0 & - & 0 & + & \text{pole} & - & \text{pole} & + \end{array}$

---

- Step 2. Structure of $f(x)$ **

According to the given conditions, we can express $f(x) = \frac{(x-p_1)(x-p_2)(x-p_3)}{(x-q_1)(x-q_2)(x-q_3)}.$ Combining this with the sign analysis, we can write $f(x) = \frac{(x-a)(x-b)(x-p_3)}{(x-c)(x-d)(x-q_3)}.$ We can separate this into two factors: $f(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)} \cdot \frac{x-p_3}{x-q_3}.$ Define $f_1(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)}, \qquad f_2(x) = \frac{x-p_3}{x-q_3}.$

---

- Step 3. Determining $p_3$ and $q_3$ **

By analyzing the sign changes of $f_1(x)$ , we see that it already matches the required sign pattern for $f(x)$ . Therefore, $f_2(x)$ must remain positive on every interval; otherwise, it would introduce extra sign changes.

This means $p_3 = q_3$ , since if $p_3 \ne q_3$ , the factor $\frac{x-p_3}{x-q_3}$ would change sign between $p_3$ and $q_3$ . Let $p_3 = q_3 = t$ .

Furthermore: - $t$ cannot lie within $[a,b] \cup (c,d)$ , since that would alter the set $\{x : f(x) \le 0\}$ . - $t$ cannot be equal to $a$ or $b$ , since that would make those endpoints holes (discontinuities), contradicting that $[a,b]$ is a *closed* interval.

Hence, $t$ must either be equal to $c$ or $d$ , or lie outside $[a,b] \cup (c,d)$ .

---

- Step 4. Possible values of $a,b,c,d,t$ **

Given $a<b<c<d$ and $\{x : f(x) \le 0\} = [a,b] \cup (c,d)$ , we choose four distinct numbers from $\{1,2,3,4,5\}$ for $a,b,c,d$ : $\binom{5}{4} = 5.$ The five possible cases are: $[1,2]\cup(3,4), \quad [1,2]\cup(3,5), \quad [1,2]\cup(4,5), \quad [1,3]\cup(4,5), \quad [2,3]\cup(4,5).$

---

- Case 1:** $[1,2]\cup(3,4)$

$t$ can be $3$ , $4$ , or $5$ , giving 3 functions: $f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-4)(x-3)}, \quad f(x) = \frac{(x-1)(x-2)(x-4)}{(x-3)(x-4)(x-4)}, \quad f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-4)(x-5)}.$

- Case 2:** $[1,2]\cup(3,5)$

$t$ can be $3$ or $5$ , giving 2 functions: $f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-5)(x-3)}, \quad f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-5)(x-5)}.$

- Case 3:** $[1,2]\cup(4,5)$

$t$ can be $3$ , $4$ , or $5$ , giving 3 functions: $f(x) = \frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-3)}, \quad f(x) = \frac{(x-1)(x-2)(x-4)}{(x-4)(x-5)(x-4)}, \quad f(x) = \frac{(x-1)(x-2)(x-5)}{(x-4)(x-5)(x-5)}.$

- Case 4:** $[1,3]\cup(4,5)$

$t$ can be $4$ or $5$ , giving 2 functions: $f(x) = \frac{(x-1)(x-3)(x-4)}{(x-4)(x-5)(x-4)}, \quad f(x) = \frac{(x-1)(x-3)(x-5)}{(x-4)(x-5)(x-5)}.$

- Case 5:** $[2,3]\cup(4,5)$

$t$ can be $1$ , $4$ , or $5$ , giving 3 functions: $f(x) = \frac{(x-2)(x-3)(x-1)}{(x-4)(x-5)(x-1)}, \quad f(x) = \frac{(x-2)(x-3)(x-4)}{(x-4)(x-5)(x-4)}, \quad f(x) = \frac{(x-2)(x-3)(x-5)}{(x-4)(x-5)(x-5)}.$

---

- Step 5. Final Answer**

Summing all possible cases: $3 + 2 + 3 + 2 + 3 = 13.$ Thus, there are $\boxed{13}$ possible functions $f(x)$ , corresponding to choice $\boxed{E}$ .

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2025 AMC 12A Problems/Problem 25: Difference between revisions

Revision as of 02:24, 7 November 2025

@@ Line 1: / Line 1: @@
-Polynomials <imath>P(x)</imath> and <imath>Q(x)</imath> each have degree <imath>3</imath> and leading coefficient <imath>1</imath>, and their roots are all elements of <imath>\{1,2,3,4,5\}</imath>. The function <imath>f(x) = \tfrac{P(x)}{Q(x)}</imath> has the property that there exist real numbers <imath>a < b < c < d</imath> such that the set of all real numbers <imath>x</imath> such that <imath>f(x) \leq 0</imath> consists of the closed interval <imath>[a,b]</imath> together with the open interval <imath>(c,d)</imath>. How many functions <imath>f(x)</imath> are possible?
+We begin by analyzing the sign chart. From the condition <imath>f(x) \le 0</imath> on <imath>[a,b] \cup (c,d)</imath> and <imath>f(x) > 0</imath> elsewhere, we deduce the following:
-==Solution 1 by Victor Zhang ==
+- <imath>a</imath> and <imath>b</imath> are zeros of <imath>f</imath>, since we transition from <imath>f>0</imath> to <imath>f\le0</imath> at <imath>a</imath>, and from <imath>f\le0</imath> to <imath>f>0</imath> at <imath>b</imath>.
-\documentclass[12pt]{article}
+- <imath>c</imath> and <imath>d</imath> are poles or holes of <imath>f</imath>, since on <imath>(c,d)</imath> we have <imath>f\le0</imath>, while immediately outside that interval <imath>f>0</imath>.
-\usepackage{amsmath, amssymb, amsthm}
-\usepackage{enumitem}
-\begin{document}
-\section*{Solution}
-\subsection*{Step 1: Sign chart analysis}
+Thus, the sign pattern of <imath>f(x)</imath> is
-From the structure <imath>f(x) \le 0</imath> on <imath>[a,b] \cup (c,d)</imath> and <imath>f(x) &gt; 0</imath> elsewhere,
+<cmath>
-we deduce:
-\begin{itemize}
-\item <imath>a</imath> and <imath>b</imath> are zeros of <imath>f</imath> (since we transition from <imath>f &gt; 0</imath> to <imath>f \le
-</imath> at <imath>a</imath> and from <imath>f \le 0</imath> to <imath>f &gt; 0</imath> at <imath>b</imath>).
-\item <imath>c</imath> and <imath>d</imath> are poles or holes of <imath>f</imath> (since on <imath>(c,d)</imath> we have <imath>f \le
-</imath>, at <imath>c^+</imath> and <imath>d^-</imath> the sign is negative, and immediately outside the
-interval <imath>f &gt; 0</imath>).
-\end{itemize}
-Thus the sign pattern is:\\
-<imath>\textstyle
 \begin{array}{cccccccccc}
-&amp; (-\infty, a) &amp; a &amp; (a,b) &amp; b &amp; (b,c) &amp; c &amp; (c,d) &amp; d &amp; (d,\infty) \\
+ & (-\infty, a) & a & (a,b) & b & (b,c) & c & (c,d) & d & (d,\infty) \\
-f(x) &amp; + &amp; 0 &amp; - &amp; 0 &amp; + &amp; \text{pole} &amp; - &amp; \text{pole} &amp; +
+f(x) & + & 0 & - & 0 & + & \text{pole} & - & \text{pole} & +
 \end{array}
-</imath>
+</cmath>
-\section*{Step 2: Structure of \( f(x) \)}
-According to the given conditions, \( f(x) \) can be expressed as:
+---
-\[
+**Step 2. Structure of <imath>f(x)</imath>**
+According to the given conditions, we can express
+<cmath>
 f(x) = \frac{(x-p_1)(x-p_2)(x-p_3)}{(x-q_1)(x-q_2)(x-q_3)}.
-\]
+</cmath>
-Combining the sign analysis from Step 1, we can rewrite this expression as:
+Combining this with the sign analysis, we can write
-\[
+<cmath>
 f(x) = \frac{(x-a)(x-b)(x-p_3)}{(x-c)(x-d)(x-q_3)}.
-\]
+</cmath>
-Furthermore, we can break down the expression into two parts:
+We can separate this into two factors:
-\[
+<cmath>
 f(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)} \cdot \frac{x-p_3}{x-q_3}.
-\]
+</cmath>
-Defining \( f(x) = f_1(x) \cdot f_2(x) \), we have:
+Define
-\[
+<cmath>
-f_1(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)}, \quad f_2(x) = \frac{x-p_3}{x-q_3}.
+f_1(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)}, \qquad f_2(x) = \frac{x-p_3}{x-q_3}.
-\]
+</cmath>
-\section*{Step 3: Analysis of \( p_3 \) and \( q_3 \)}
+---
-\begin{itemize}
-\item By analyzing the sign changes of \( f_1(x) \), we find that it completely
-satisfies the sign chart for \( f(x) \). We can therefore conclude that \( f_2(x) \)
-must be positive on every interval; otherwise, it would introduce an extra sign
-change. This forces the condition \( p_3 = q_3 \), because if \( p_3 \neq q_3 \),
-the factor \( f_2(x) \) would be negative on the interval between \( p_3 \) and \(
-q_3 \). Let \( p_3 = q_3 = t \).
-\item Furthermore, \( t \) cannot lie within \( [a,b] \cup (c,d) \). If it did, it
-would contradict the given condition that the set \( \{x : f(x) \le 0\} \) consists of
-the closed interval \( [a, b] \) and the open interval \( (c, d) \).
-\item Additionally, \( t \) cannot be equal to \( a \) or \( b \). If it were, that
-point would be a hole (a point of discontinuity) in the graph of \( f(x) \), which
-contradicts the fact that \( [a, b] \) is a \textbf{closed} interval where \( f(x) \le 0
-\). On a closed interval, the function must be defined at the endpoints.
-\end{itemize}
-In conclusion, \( t \) can only be equal to \( c \) or \( d \), or lie in an interval
-outside of \( [a,b] \cup (c,d) \).
-\section*{Step 4: The Possible Values of <imath>a</imath>, <imath>b</imath>, <imath>c</imath>, <imath>d</imath>, and <imath>t</imath>}
-\begin{itemize}
-\item Given <imath>a &lt; b &lt; c &lt; d</imath> and the condition that <imath>\{x: f(x) \leq 0\} = [a,b]
-\cup (c,d)</imath>, the number of ways to choose <imath>a,b,c,d</imath> from <imath>\{1,2,3,4,5\}</imath> is:
-\[
-\binom{5}{4} = 5
-\]
-The five cases are:
-<imath>[1,2]\cup (3,4)</imath>; <imath>[1,2]\cup (3,5)</imath>; <imath>[1,2]\cup (4,5)</imath>; <imath>[1,3]\cup (4,5)</imath>;
-<imath>[2,3]\cup (4,5)</imath>
-\item \textbf{Case 1:} <imath>[1,2]\cup (3,4)</imath>, <imath>t</imath> can be 3, 4, or 5.\\
-This gives us 3 combinations:
-\[
-f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-4)(x-3)}
-\]
-\[
-f(x) = \frac{(x-1)(x-2)(x-4)}{(x-3)(x-4)(x-4)}
-\]
-\[
-f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-4)(x-5)}
-\]
-\item \textbf{Case 2:} <imath>[1,2]\cup (3,5)</imath>, <imath>t</imath> can be 3 or 5.\\
-This gives us 2 combinations:
+**Step 3. Determining <imath>p_3</imath> and <imath>q_3</imath>**
-\[
-f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-5)(x-3)}
-\]
-\[
-f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-5)(x-5)}
-\]
-\item \textbf{Case 3:} <imath>[1,2]\cup (4,5)</imath>, <imath>t</imath> can be 3, 4, or 5.\\
-This gives us 3 combinations:
-\[
-f(x) = \frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-3)}
-\]
-\[
-f(x) = \frac{(x-1)(x-2)(x-4)}{(x-4)(x-5)(x-4)}
-\]
-\[
-f(x) = \frac{(x-1)(x-2)(x-5)}{(x-4)(x-5)(x-5)}
-\]
-\item \textbf{Case 4:} <imath>[1,3]\cup (4,5)</imath>, <imath>t</imath> can be 4 or 5.\\
-This gives us 2 combinations:
-\[
-f(x) = \frac{(x-1)(x-3)(x-4)}{(x-4)(x-5)(x-4)}
-\]
-\[
-f(x) = \frac{(x-1)(x-3)(x-5)}{(x-4)(x-5)(x-5)}
-\]
-\item \textbf{Case 5:} <imath>[2,3]\cup (4,5)</imath>, <imath>t</imath> can be 1, 4, or 5.\\
-This gives us 3 combinations:
-\[
-f(x) = \frac{(x-2)(x-3)(x-1)}{(x-4)(x-5)(x-1)}
-\]
-\[
-f(x) = \frac{(x-2)(x-3)(x-4)}{(x-4)(x-5)(x-4)}
-\]
-\[
-f(x) = \frac{(x-2)(x-3)(x-5)}{(x-4)(x-5)(x-5)}
-\]
-\end{itemize}
-\section*{Step 5: Final Answer}
-Summing up all possible cases from Step 4 gives us <imath>3 + 2 + 3 + 2 + 3 = 13</imath>
-possible functions <imath>f(x)</imath>.
-Thus the answer is <imath>\boxed{E}</imath>.
-\end{document}
-\section*{Step 4: The possible of a,b,c,d, and t }
+By analyzing the sign changes of <imath>f_1(x)</imath>, we see that it already matches the required sign pattern for <imath>f(x)</imath>. Therefore, <imath>f_2(x)</imath> must remain positive on every interval; otherwise, it would introduce extra sign changes.
-\begin{itemize}
-\item According to given <imath>a &lt; b &lt; c &lt; d </imath> and \( [a,b] \cup (c,d) \), the ways
-of choose a,b,c,d from {1,2,3,4,5}:
-<cmath> \binom{5}{4}=5</cmath>
-List 5 cases as:\\$[1,2]\cup (3,4); [1,2]\cup (3,5); [1,2]\cup (4,5); [1,3]\cup
-(4,5); [2,3]\cup (4,5)<imath>
-\item Case 1: </imath>[1,2]\cup (3,4)<imath>, t can be 3,4, or 5.\\
+This means <imath>p_3 = q_3</imath>, since if <imath>p_3 \ne q_3</imath>, the factor <imath>\frac{x-p_3}{x-q_3}</imath> would change sign between <imath>p_3</imath> and <imath>q_3</imath>.
-Give us 3 combnation:
+Let <imath>p_3 = q_3 = t</imath>.
-\[
-f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-4)(x-3)}
+Furthermore:
-\]
+- <imath>t</imath> cannot lie within <imath>[a,b] \cup (c,d)</imath>, since that would alter the set <imath>\{x : f(x) \le 0\}</imath>.
-\[
+- <imath>t</imath> cannot be equal to <imath>a</imath> or <imath>b</imath>, since that would make those endpoints holes (discontinuities), contradicting that <imath>[a,b]</imath> is a *closed* interval.
-f(x) = \frac{(x-1)(x-2)(x-4)}{(x-3)(x-4)(x-4
-)}
+Hence, <imath>t</imath> must either be equal to <imath>c</imath> or <imath>d</imath>, or lie outside <imath>[a,b] \cup (c,d)</imath>.
-\]
-\[
+---
+**Step 4. Possible values of <imath>a,b,c,d,t</imath>**
+Given <imath>a<b<c<d</imath> and <imath>\{x : f(x) \le 0\} = [a,b] \cup (c,d)</imath>, we choose four distinct numbers from <imath>\{1,2,3,4,5\}</imath> for <imath>a,b,c,d</imath>:
+<cmath>
+\binom{5}{4} = 5.
+</cmath>
+The five possible cases are:
+<cmath>
+[1,2]\cup(3,4), \quad [1,2]\cup(3,5), \quad [1,2]\cup(4,5), \quad [1,3]\cup(4,5), \quad [2,3]\cup(4,5).
+</cmath>
+---
+**Case 1:** <imath>[1,2]\cup(3,4)</imath>
+<imath>t</imath> can be <imath>3</imath>, <imath>4</imath>, or <imath>5</imath>, giving 3 functions:
+<cmath>
+f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-4)(x-3)}, \quad
+f(x) = \frac{(x-1)(x-2)(x-4)}{(x-3)(x-4)(x-4)}, \quad
 f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-4)(x-5)}.
-\]
+</cmath>
-\item Case 2: </imath>[1,2]\cup (3,5)<imath>, t can be 3, or 5.\\
-Give us 2 combnation:
+**Case 2:** <imath>[1,2]\cup(3,5)</imath>
-\[
+<imath>t</imath> can be <imath>3</imath> or <imath>5</imath>, giving 2 functions:
-f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-5)(x-3)}
+<cmath>
-\]
+f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-5)(x-3)}, \quad
-\[
+f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-5)(x-5)}.
-f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-4)(x-5
+</cmath>
-)}
-\]
+**Case 3:** <imath>[1,2]\cup(4,5)</imath>
-\item Case 3: </imath>[1,2]\cup (4,5)<imath>, t can be 3,4, or 5.\\
+<imath>t</imath> can be <imath>3</imath>, <imath>4</imath>, or <imath>5</imath>, giving 3 functions:
-Give us 3 combnation:
+<cmath>
-\[
+f(x) = \frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-3)}, \quad
-f(x) = \frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-3)}
+f(x) = \frac{(x-1)(x-2)(x-4)}{(x-4)(x-5)(x-4)}, \quad
-\]
-\[
-f(x) = \frac{(x-1)(x-2)(x-4)}{(x-4)(x-5)(x-4
-)}
-\]
-\[
 f(x) = \frac{(x-1)(x-2)(x-5)}{(x-4)(x-5)(x-5)}.
-\]
+</cmath>
-\item Case 4: </imath>[1,3]\cup (4,5)<imath>, t can be 4, or 5.\\
-Give us 2 combnation:
+**Case 4:** <imath>[1,3]\cup(4,5)</imath>
-\[
+<imath>t</imath> can be <imath>4</imath> or <imath>5</imath>, giving 2 functions:
-f(x) = \frac{(x-1)(x-3)(x-4)}{(x-4)(x-5)(x-4)}
+<cmath>
+f(x) = \frac{(x-1)(x-3)(x-4)}{(x-4)(x-5)(x-4)}, \quad
+f(x) = \frac{(x-1)(x-3)(x-5)}{(x-4)(x-5)(x-5)}.
+</cmath>
-\]
+**Case 5:** <imath>[2,3]\cup(4,5)</imath>
-\[
+<imath>t</imath> can be <imath>1</imath>, <imath>4</imath>, or <imath>5</imath>, giving 3 functions:
-f(x) = \frac{(x-1)(x-3)(x-5)}{(x-4)(x-5)(x-5
+<cmath>
-)}
+f(x) = \frac{(x-2)(x-3)(x-1)}{(x-4)(x-5)(x-1)}, \quad
-\]
+f(x) = \frac{(x-2)(x-3)(x-4)}{(x-4)(x-5)(x-4)}, \quad
-\item Case 5: </imath>[2,3]\cup (4,5)<imath>, t can be 1,4, or 5.\\
+f(x) = \frac{(x-2)(x-3)(x-5)}{(x-4)(x-5)(x-5)}.
-Give us 3 combnation:
+</cmath>
-\[
-f(x) = \frac{(x-2)(x-3)(x-1)}{(x-4)(x-5)(x-1)}
-\]
-\[
-f(x) = \frac{(x-2)(x-3)(x-4)}{(x-4)(x-5)(x-4)}
-\] \[
-f(x) = \frac{(x-2)(x-3)(x-5)}{(x-4)(x-5)(x-5)}
-\]
-\section*{Step 5: final answer }
+---
-\Sum up all possible cases of step 4 give us 13 possible functions f(x).
-Thus the answer is </imath>\boxed{E}$.
+**Step 5. Final Answer**
-\end{document}
-==Video Solution 1 by OmegaLearn==
+Summing all possible cases:
-https://youtu.be/SPbTyq3Dz_0
+<cmath>
++ 2 + 3 + 2 + 3 = 13.
+</cmath>
+Thus, there are <imath>\boxed{13}</imath> possible functions <imath>f(x)</imath>, corresponding to choice <imath>\boxed{E}</imath>.